# Does the series converge $\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$

Does the series converge? $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$$

I tried to solve it with alternating series test and got $$\lim_{n\rightarrow\infty}a_n=1\neq0\implies\text{??Divergence?? or how to prove that}$$

Idea: To show that the limit of $$(-1)^n$$ doesnt exist, thus the series is divergent. Is this correct?

You're correct, this series is not convergent because sequence $$\frac{(-1)^n}{\cos\frac{\pi}{2n}}$$ is not convergent. For a series $$\sum_n {a_n}$$ to be convergent, it is necessary (but not yet sufficient) that $$\lim_{n\to\infty} a_n$$ exists and is equal to $$0$$.

• and because this limit doesn't tend to zero it is sufficient to say that the series diverges?
– VLC
Feb 27, 2021 at 21:08
• @BiliDebili the vanishing condition states that if a series is summable then $lim_{n \to \infty} a_n =0$. Since the series in question doesn't (the consequent, the latter condition) then by modus tollens the series is not summable. Feb 27, 2021 at 21:12
• This sequence doesn't have a limit, in particular it doesn't tend to zero, which is sufficient to tell that the series does not converge. Feb 27, 2021 at 21:18

Series diverges, as general member doesn't tends to zero.

Think about the behavior of $$\cos\frac{\pi}{2n}$$ for many $$n$$'s. Then evaluate the partial sums as $$n$$ increases.

It's a bit of a trick question.